Calculate the half life of the first-order reaction:
`C_(2)H_(4)O(g) rarr CH_(4)(g)+CO(g)`
The initial pressure of `C_(2)H_(4)O(g)` is `80 mm` and the total pressure at the end of `20 min` is `120 mm`.

Options

A. `40 min`
B. `120 min`
C. `20 min`
D. `80 min`

Explanation

To calculate the half-life of the first-order reaction given by the equation:

\[ C_{2}H_{4}O(g) \rightarrow CH_{4}(g) + CO(g) \]

we will follow these steps:

### Step 1: Identify the initial and final conditions
- Initial pressure of \( C_{2}H_{4}O(g) \) = 80 mm
- Total pressure at the end of 20 minutes = 120 mm

### Step 2: Calculate the change in pressure
The change in pressure due to the reaction can be calculated as follows:
- Change in pressure = Total pressure at the end - Initial pressure of \( C_{2}H_{4}O(g) \)
- Change in pressure = 120 mm - 80 mm = 40 mm

### Step 3: Determine the remaining pressure of \( C_{2}H_{4}O(g) \)
- Remaining pressure of \( C_{2}H_{4}O(g) \) = Initial pressure - Change in pressure
- Remaining pressure = 80 mm - 40 mm = 40 mm

### Step 4: Use the first-order rate equation to find the rate constant \( K \)
For a first-order reaction, the rate constant \( K \) can be calculated using the formula:

\[ K = \frac{2.303}{T} \log\left(\frac{[C_{2}H_{4}O]_{initial}}{[C_{2}H_{4}O]_{remaining}}\right) \]

Where:
- \( T = 20 \) minutes
- Initial concentration = 80 mm
- Remaining concentration = 40 mm

Plugging in the values:

\[ K = \frac{2.303}{20} \log\left(\frac{80}{40}\right) \]

### Step 5: Calculate the logarithm
- \( \log\left(\frac{80}{40}\right) = \log(2) \)
- The value of \( \log(2) \approx 0.3010 \)

### Step 6: Substitute and calculate \( K \)
\[ K = \frac{2.303}{20} \times 0.3010 \]

Calculating this gives:

\[ K \approx \frac{2.303 \times 0.3010}{20} \approx 0.0346 \text{ min}^{-1} \]

### Step 7: Calculate the half-life \( T_{1/2} \)
The half-life for a first-order reaction is given by the formula:

\[ T_{1/2} = \frac{0.693}{K} \]

Substituting the value of \( K \):

\[ T_{1/2} = \frac{0.693}{0.0346} \]

Calculating this gives:

\[ T_{1/2} \approx 20 \text{ minutes} \]

### Final Answer
The half-life of the reaction is **20 minutes**.

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